Let Sl := {x e IR: x ::: OJ;Show in detail that the set Sl has lower bounds, but no

Step 1 of 2

Let \(S_{1}=\{x \in \mathbb{R}: x \geq 0\}\)

To prove that \(S_{1}\) has lower bounds, but no upper bounds.

By definition a set S is called bounded below if there exist a real number b such that \(x \geq b\) for all \(x \in S\). The real number b is the lower bound of S.

From the definition of the set \(S_1\), it is seen that for all \(x \in S, x \geq 0\). Therefore, \(S_{1}\) is bounded below and \(0\) is a lower bound of the set \(S_1\).

Let us consider an arbitrary element \(n \in S_{1}\). Then, \(n \geq 0\).

Now, \((1+n)>0\). Therefore, \((1+n) \in S_{1}\) and \(n<(1+n)\).

Since n is chosen arbitrarily, for every element \(n \in S_{1}\), we always find a number greater than n which is also an element of \(S_{1}\). As a result, we can’t find any real number B such that \(n \leq B\) for all \(n \in S_{1}\). It shows that \(S_{1}\) has no upper bounds.